Log Equation

Jul 2009
36
0
Hello, Could you help me solve this ?




I'd appreciate ur help, thanks a lot
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Hello, Could you help me solve this ?




I'd appreciate ur help, thanks a lot
\(\displaystyle \frac{\log_2{(x^2 + 2x - 7)}}{\log_2{(x - 3)}} = 1\)

\(\displaystyle \log_2{(x^2 + 2x - 7)} = \log_2{(x - 3)}\)

\(\displaystyle x^2 + 2x - 7 = x - 3\)

\(\displaystyle x^2 + x - 4 = 0\)

\(\displaystyle x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 4 = 0\)

\(\displaystyle \left(x + \frac{1}{2}\right)^2 - \frac{17}{4} = 0\)

\(\displaystyle \left(x + \frac{1}{2}\right)^2 = \frac{17}{4}\)

\(\displaystyle x + \frac{1}{2}= \pm \frac{\sqrt{17}}{2}\)

\(\displaystyle x = -\frac{1}{2} \pm \frac{\sqrt{17}}{2}\).



Therefore

\(\displaystyle x = \frac{-1 - \sqrt{17}}{2}\) or \(\displaystyle x = \frac{-1 + \sqrt{17}}{2}\).
 
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